Chapter 8: Hypothesis Testing with One Sample
Chapter 8 Homework
8.1 Homework
Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, [latex]H_{0}[/latex], and the alternative hypothesis. [latex]H_{a}[/latex], in terms of the appropriate parameter ([latex]\mu[/latex] or [latex]p[/latex]).
- The mean number of years Americans work before retiring is 34.
- At most 60% of Americans vote in presidential elections.
- The mean starting salary for San Jose State University graduates is at least $100,000 per year.
- Twenty-nine percent of high school seniors get drunk each month.
- Fewer than 5% of adults ride the bus to work in Los Angeles.
- The mean number of cars a person owns in her lifetime is not more than ten.
- About half of Americans prefer to live away from cities, given the choice.
- Europeans have a mean paid vacation each year of six weeks.
- The chance of developing breast cancer is under 11% for women.
- Private universities’ mean tuition cost is more than $20,000 per year.
Solution
- [latex]H_{0}: \mu = 34; H_{a}: \mu \neq 34[/latex]
- [latex]H_{0}: p \le 0.60; H_{a}: p > 0.60[/latex]
- [latex]H_{0}: \mu \ge 100,000; H_{a}: \mu \lt 100,000[/latex]
- [latex]H_{0}: p = 0.29; H_{a}: p \neq 0.29[/latex]
- [latex]H_{0}: p = 0.05; H_{a}: p \lt 0.05[/latex]
- [latex]H_{0}: \mu \le 10; H_{a}: \mu > 10[/latex]
- [latex]H_{0}: p = 0.50; H_{a}: p \neq 0.50[/latex]
- [latex]H_{0}: \mu= 6; H_{a}: \mu \neq 6[/latex]
- [latex]H_{0}: p \ge 0.11; H_{a}: p \lt 0.11[/latex]
- [latex]H_{0}: \mu \le 20,000; H_{a}: \mu > 20,000[/latex]
Over the past few decades, public health officials have examined the link between weight concerns and teen girls’ smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:
- [latex]p \lt 0.30[/latex]
- [latex]p \le 0.30[/latex]
- [latex]p \ge 0.30[/latex]
- [latex]p > 0.30[/latex]
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:
a. [latex]p = 0.20[/latex]
b. [latex]p > 0.20[/latex]
c. [latex]p \lt 0.20[/latex]
d. [latex]p \le 0.20[/latex]
Solution
c
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:
a. [latex]H_{0}: \overline{x} = 4.5, H_{a}: \overline{x} > 4.5[/latex]
b. [latex]H_{0}: \mu \ge 4.5, H_{a}: \mu \lt 4.5[/latex]
c. [latex]H_{0}: \mu = 4.75, H_{a}: \mu > 4.75[/latex]
d. [latex]H_{0}: \mu = 4.5, H_{a}: \mu > 4.5[/latex]
8.2 Homework
1. State the Type I and Type II errors in complete sentences given the following statements.
- The mean number of years Americans work before retiring is 34.
- At most 60% of Americans vote in presidential elections.
- The mean starting salary for San Jose State University graduates is at least $100,000 per year.
- Twenty-nine percent of high school seniors get drunk each month.
- Fewer than 5% of adults ride the bus to work in Los Angeles.
- The mean number of cars a person owns in his or her lifetime is not more than ten.
- About half of Americans prefer to live away from cities, given the choice.
- Europeans have a mean paid vacation each year of six weeks.
- The chance of developing breast cancer is under 11% for women.
- Private universities mean tuition cost is more than $20,000 per year.
Solution
- Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
- Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
- Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
- Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
- Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer than 5% do.
- Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
- Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
- Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
- Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
- Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.
2. For the ten statements above, answer the following in complete sentences.
- State a consequence of committing a Type I error.
- State a consequence of committing a Type II error.
When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?
a. To conclude the drug is safe when, in fact, it is unsafe.
b. Not to conclude the drug is safe when, in fact, it is safe.
c. To conclude the drug is safe when, in fact, it is safe.
d. Not to conclude the drug is unsafe when, in fact, it is unsafe.
Solution
b
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is [latex]\underline{\hspace{2cm}}[/latex].
a. at least 20%, when in fact, it is less than 20%.
b. 20%, when in fact, it is 20%.
c. less than 20%, when in fact, it is at least 20%.
d. less than 20%, when in fact, it is less than 20%.
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?
The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours
a. is more than seven hours.
b. is at most seven hours.
c. is at least seven hours.
d. is less than seven hours.
Solution
d
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:
a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher
8.3 Homework
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is [latex]\overline{X} \sim \underline{\hspace{2cm}}[/latex].
a. [latex]N\left(7.24,\frac{1.93}{\sqrt{22}}\right)[/latex]
b. [latex]N\left(7.24,1.93\right)[/latex]
c. [latex]t_{22}[/latex]
d. [latex]t_{21}[/latex]
Solution
d
8.4 Homework
The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.
- Is this a test of one mean or proportion?
- State the null and alternative hypotheses.
- Is this a right-tailed, left-tailed, or two-tailed test?
- What symbol represents the random variable for this test?
- In words, define the random variable for this test.
- Calculate the following:
- [latex]x = \underline{\hspace{2cm}}[/latex]
- [latex]n = \underline{\hspace{2cm}}[/latex]
- [latex]{p}^{\prime } = \underline{\hspace{2cm}}[/latex]
- Calculate [latex]\sigma_{x} = \underline{\hspace{2cm}}[/latex]. Show the formula set-up.
- State the distribution to use for the hypothesis test.
- Find the p-value.
- At a pre-conceived [latex]\alpha = 0.05[/latex], what is your:
- Decision:
- Reason for the decision:
- Conclusion (write out in a complete sentence):
8.5 Homework
For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link]. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
Note
If you are using a Student’s-t distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using [latex]\alpha = 0.05[/latex], is the data highly inconsistent with the claim?
Solution
a. [latex]H_0: \mu \ge 50,000[/latex]
b. [latex]H_a: \mu \lt 50,000[/latex]
c. Let [latex]\overline{X} =[/latex] the average lifespan of a brand of tires.
d. normal distribution
e. [latex]z = -2.315[/latex]
f. [latex]\text{p-value} = 0.0103[/latex]
g. Check student’s solution.
h. i. Alpha: 0.05
ii. Decision: Reject the null hypothesis.
iii. Reason for decision: The p-value is less than 0.05.
iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
i. (43537, 49463)
From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?
Solution
a. [latex]H_0: \mu = \$1.00[/latex]
b. [latex]H_a: \mu \neq \$1.00[/latex]
c. Let [latex]\overline{X} =[/latex] the average cost of a daily newspaper.
d. normal distribution
e. [latex]z = -0.866[/latex]
f. [latex]\text{p-value} = 0.3865[/latex]
g. Check student’s solution.
h. i. Alpha: 0.01
ii. Decision: Do not reject the null hypothesis.
iii. Reason for decision: The p-value is greater than 0.01.
iv. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
i. ($0.84, $1.06)
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?
Solution
a. [latex]H_0: \mu = 10[/latex]
b. [latex]H_a: \mu \neq 10[/latex]
c. Let [latex]\overline{X} =[/latex] the mean number of sick days an employee takes per year.
d. Student’s t-distribution
e. [latex]t = -1.12[/latex]
f. [latex]\text{p-value} = 0.300[/latex]
g. Check student’s solution.
h. i. Alpha: 0.05
ii. Decision: Do not reject the null hypothesis.
iii. Reason for decision: The p-value is greater than 0.05.
iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
i. (4.9443, 11.806)
In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women’s movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?
Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can’t quite figure out, most people don’t believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?
Solution
a. [latex]H_0: p \ge 0.6[/latex]
b. [latex]H_a: p \lt 0.6[/latex]
c. Let [latex]P^{\prime} =[/latex] the proportion of students who feel more enriched as a result of taking Elementary Statistics.
d. normal for a single proportion
e. 1.12
f. [latex]\text{p-value} = 0.1308[/latex]
g. Check student’s solution.
h. i. Alpha: 0.05
ii. Decision: Do not reject the null hypothesis.
iii. Reason for decision: The p-value is greater than 0.05.
iv. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.
i. Confidence Interval: (0.409, 0.654)
The “plus-4s” confidence interval is (0.411, 0.648)
1. A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.
2. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.
Solution to part 2
a. [latex]H_0: \mu = 4[/latex]
b. [latex]H_a: \mu \neq 4[/latex]
c. Let [latex]\overline{X} =[/latex] the average I.Q. of a set of brown trout.
d. two-tailed Student’s t-test
e. [latex]t = 1.95[/latex]
f. [latex]\text{p-value} = 0.076[/latex]
g. Check student’s solution.
h. i. Alpha: 0.05
ii. Decision: Reject the null hypothesis.
iii. Reason for decision: The p-value is greater than 0.05
iv. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
i. (3.8865, 5.9468)
According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percentage of girls born in China is 46.7?
A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percentage is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.
Solution
a. [latex]H_0: p \ge 0.13[/latex]
b. [latex]H_a: p \lt 0.13[/latex]
c. Let [latex]P^{\prime} =[/latex] the proportion of Americans who have seen or sensed angels
d. normal for a single proportion
e. –2.688
f. [latex]\text{p-value} = 0.0036[/latex]
g. Check student’s solution.
h. i. Alpha: 0.05
ii. Decision: Reject the null hypothesis.
iii. Reason for decision: The p-value is less than 0.05.
iv. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.
i. (0, 0.0623).
The “plus-4s” confidence interval is (0.0022, 0.0978)
The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?
Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
